Answer:
The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Explanation:
Given;
average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²
distance traveled by the commercial Jet liner, s = 1542 m
The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the commercial Jet liner = 0
s = 0 + ¹/₂at²
s = ¹/₂at²
2s = at²
Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Answer:
a) t = 3.027 10⁻⁹ s
, b) y = 2.25 10⁻² m
Explanation:
We can solve this problem using the kinematic relations
a) as on the x-axis there is no relationship
vₓ = x / t
t = x / vₓ
We reduce the magnitudes to the SI system
x = 5.6 cm (1m / 100 vm) = 0.056 m
we calculate
t = 0.056 / 1.85 10⁷
t = 3.027 10⁻⁹ s
b) the time is the same for the two movements, on the y axis
y = v₀t + ½ a t²
as the beam leaves horizontal there is no initial vertical velocity
y = ½ a t²
let's calculate
y = ½ 5.45 10¹⁵ (3.027 10⁻⁹)²
y = 2.25 10⁻² m
Sum of the first ten digits is:
7+3+4+5+4+1+7+8+0 = 39
39, when divided by 9 give you the remainder of 3
9 x4, is 36
36 +3 equals 39
(39 = 9 x 4 + 3)
So X= 0
Well most of the earth's oxygen resides in mineral oxides of the crust. a small fraction resides in the atmosphere and an even smaller fraction resides in the biosphere. still the biosphere is crucial to understanding the atmospheric oxygen budget as it controls short term exchanges between sediments and the atmosphere.
